scott wrote:
>> If we select K balls, the probability of ball #1 appearing becomes 1/N 
>> + 1/N + 1/N + 1/N + ... K times over, minus the probability of ball #1 
>> appearing more than once.
>>
>> In other words, we have K/N - E, and I can't figure out how to compute E.
> 
> You can think of it the other way round, ie the probability of ball #1 
> appearing after K tries is 1 minus the probability of ball #1 not 
> occurring after K tries.

Ah. So instead of a chain of non-exclusive OR events, you construct a 
chain of independent AND events? Ingenius...

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